Dobrushin's contraction for positive stochastic matrices

Step 1: rows are close. Let $p,q$ be two rows of $P$, probability vectors with all entries $\ge\delta$. The vector $d=p-q$ has zero sum, so $\sum_id_i^+=\sum_id_i^-=\frac12\|d\|_1$, where $d^\pm$ are the positive/negative parts. Moreover
$$\sum_id_i^+=\sum_i(p_i-q_i)^+=1-\sum_i\min(p_i,q_i)\le1-n\delta.$$

Step 2: oscillation contracts. For $x\in\mathbb{R}^n$, $(Px)_r-(Px)_s=d\cdot x$ with $d=p_r-p_s$. Because $\sum d_i=0$,
$$d\cdot x=\sum d_i^+x_i-\sum d_i^-x_i\le\Big(\sum d_i^+\Big)\max x-\Big(\sum d_i^-\Big)\min x=\Big(\sum d_i^+\Big)\operatorname{osc}(x).$$
Hence $\operatorname{osc}(Px)\le(1-n\delta)\operatorname{osc}(x)$, and by induction $\operatorname{osc}(P^kx)\le(1-n\delta)^k\operatorname{osc}(x)$.

Step 3: $P^kx$ converges to a constant. Since $P$ is stochastic, each coordinate of $Px$ is an average of the coordinates of $x$: $\max P^kx$ is nonincreasing, $\min P^kx$ is nondecreasing, and both converge; as the oscillation tends to $0$, they converge to the same limit $c(x)$. So $P^kx\to c(x)\mathbf{1}$, with $|P^kx-c(x)\mathbf{1}|_\infty\le(1-n\delta)^k\operatorname{osc}(x)$.

Step 4: the limit matrix. $c$ is linear in $x$ (a limit of linear maps), so $c(x)=\pi\cdot x$ for $\pi_i=c(e_i)$. Taking $x=e_i$ with $c(e_i)\ge0$ (the entries of $P^k$ are $\ge0$) and $x=\mathbf{1}$ with $c(\mathbf{1})=1$ shows that $\pi_i\ge0$ and $\sum\pi_i=1$. Therefore $P^kx\to\mathbf{1}\pi^Tx$ for each $x$, i.e. $P^k\to\mathbf{1}\pi^T$.

Example. $P=\begin{pmatrix}0.9&0.1\\0.2&0.8\end{pmatrix}$ has $\delta=0.1$ and ratio $\le0.8$ (the actual second eigenvalue is $0.7$); $\pi=(\frac23,\frac13)$.

Remark. The constant $1-n\delta$ is the Dobrushin ergodicity coefficient. Also $\pi P=\pi$ follows from $\mathbf{1}\pi^TP=\lim P^{k+1}=\mathbf{1}\pi^T$, and $\pi$ is unique, so the Markov chain forgets its initial state exponentially fast.